The chief intent of this experiment is to utilize wind tunnel to transport out trial on aerofoil. Furthermore, this study will particular seeks to turn to the undermentioned countries ;

The theory of air current tunnel testing, this talk about the equations ( expressions ) which has been developed to used to cipher the forces which are moving on the aerofoil during the testing. It ‘s besides explains how the theory informations can be applied to work out a existent life job such as drag force and skin clash force. It is so because, by utilizing the theory consequences and understanding the forces moving upon the aerofoil. The interior decorators or applied scientists will be able to plan aerofoil for aircraft wing to cut down the retarding force and skin clash force. However, in this work you will happen how the incidence of the angle affects the lift.

The trial process describes the stairss which were used during the experiment. For illustration the first measure was to alter the angle of onslaught to -4 grade and by turning the air current tunnel machine on and leting the air to flux around aerofoil. The force per unit area which was created by the air can be so step through the manometer. The same process was used for 0 grade, 4 grade, 8 grade, 12 grade and 15 grade.

The collected information as a consequence from the trial were used to pull a figure of graphs. In the graph the force per unit area coefficient were plotted against x/c and C is the whole length of the chord and ten is the distant from the taking border to the cross subdivision countries of the aerofoil. However this clearly shows how the retarding force affects the aerofoil regard to raise. This besides shows the maximal lift or angle of onslaught. To do things simpler this work went farther and created a tabular array for analysis. Through the tabular array the information can besides evaluated for better understanding about informations obtained.

## Theory

The choice of an aerofoil for a theoretical account aircraft depends chiefly on the lift and drag features of the aerofoil. If no experimental informations are available, theoretical methods can be used to acquire an estimate of these informations. Assorted theories, premises and computations have been made by assorted academe and the professionals. Whiles other theoreticians and academia agree on some computations and premises others have differed sentiments. In some instances they used same methodological analysis but different premises and some case have wholly differed position on these affairs wholly.

Wind tunnel theory of aerodynamic features was foremost initiated by Prandtl in covering with cardinal construct in aeromechanicss. He calculated what is termed as added retarding force of an airfoil situated at a cardinal portion of a air current tunnel of a round cross-section, presuming a semi-elliptic distribution of lift ( Terazawa K and Rgakuhakusi, 1928 ) .

Glauert went farther to develop the theory even though the methodological analysis was non different from that of Prandtl. What Glauert added to the theory is that he made his computation presuming that the air current tunnel is rectangular. Both Prandtl and Glauert used the rule that: the boundaries of the cross-section are substituted by a set of images of a free whirls attach toing the airfoil at proper topographic points outside the channel so as to fulfill the boundary conditions in existent instances. In the work of Glauert, he failed to acknowledge the distance between airfoil and its nearest image.

Whereas the lift can be calculated sensible good from the frictionless pressure- severally velocity distribution on the aerofoil surface, the clash retarding force can be determined by an analysis of the boundary bed with a lesser grade of truth.

There are several methods for the design and analysis of aerofoils available. For case: profil by Professor Richard Eppler, University of Stuttgart, Germany, XFOIL by Professor Mark Drela, Massachusetts Institute of Technology, USA. However, for the interest of this work or experiment, our chief focal point was the forces which acted around the aerofoil in the air current tunnel.

Furthermore, this theory allows us to place the forces around the aerofoil and by analyzing and understanding the forces ; we could do some premises and work out how the air current would act at certain point in clip on the aerofoil. For blink of an eye, by altering the angle of incidence ( Attack ) and wind velocity in the tunnel or the velocity of the aeroplane, the air current will make force per unit areas on the aerofoil at the underside and upper surfaces at different subdivisions. This nevertheless increases the force per unit area at the underside of the aerofoil ( High force per unit area at underside ) while at the upper country, force per unit area lessenings ( Low force per unit area ) .

In world, this will do the aeroplane to raise off from the land at desire velocity. Through that, a retarding force would be created on the aerofoil. This shows that, by altering the angle, it will let the aerofoil ( aircraft ) to hold different height, force per unit area and drag force.

In other words, the angle of the onslaught can be described as an angle between the chord line and the free watercourse. It can be besides said as an angle at which the air hits the wing to make a lift. However, this shows that the sum of the lift is related to the angle of the onslaught. As the lift additions so does the angle of the onslaught additions.

The angle of the onslaught can be altering or increasing from 0 grade to 15 grades but any increment above 15 grade will do the aircraft to procrastinate. This angle can be about 17 grade. This angle can besides be called critical angle of onslaught. The diagram below depicts the angle of onslaught ( fig 1.1 ) .

V?

a

Fig 1.1

This symbol ? used to stand for the angle of onslaught and V? is free watercourse speed in the air current tunnel. However, in most instances the aerofoil ( wings ) has force per unit area distribution which causes the angle of onslaught to alter from the root to tip. Therefore, the root will hold a high angle of onslaught. The tip has a low angle of onslaught and sometime it can hold a negative angle.

In add-on, the forces which we counted at the period of the experiment are drag, dynamic viscousness, force per unit area and lift. In order to understand these forces decently, equations ( see below ) have been developed. These equations are used to find how much force will be at each given phase ( s ) , status ( s ) or job ( s ) . However, by infixing Numberss or the information collected into the equations, we can work out and visualise how the forces can impact the aerofoil.

The force per unit area on the surface of the airfoil is non unvarying. Taking aerofoil used in the experiment for case, has force per unit area distributions on the aerofoil for a given subdivision at assorted angles of onslaught. It is most appropriate to cover with non-dimensional force per unit area differences with. The force per unit area in this instance is far upstream, being used as the data point. The force per unit area coefficient is defined as:

From the diagrams below, it indicates that when the angle of onslaught is zero therefore at ( ) , it shows that the part at the olfactory organ of the aerofoil is little. Whereby at incidence angle of 6 grade the tail is positive but around most of the subdivision it becomes negative. Therefore, at the tracking edge the force per unit area coefficient comes near to +1 but this does non really make this value. Fig 1.2 shows.

Fig 1.2

This measure ?pV^2 is called dynamic force per unit area, at a low velocity this dynamic force per unit area is the same as the difference between the stagnancy force per unit area. The free watercourse force per unit area is represented by this symbol and it is equal to the dynamic and the stagnancy force per unit area. Harmonizing above statement, the equation below can be used to work out the non-dimensional force per unit area coefficient ( fig 1.3 ) .

Fig 1.3

The air current tunnel which we used to transport out the experiment ; it was an unfastened return air current tunnel. The force per unit area which was involved during proving was stagnation force per unit area and it is assumed that this force per unit area is tantamount to room force per unit area. Furthermore, the free watercourse force per unit area in this instance was step in subdivisions during the trial.

In order to acquire accuracy consequences from the trial, multi-tube manometer was used to mensurate the force per unit area differences. During the experiment we experienced that the difference in force per unit area is relative to the difference in tallness of the liquid degrees in the manometer. Through the measuring we came to cognize that the force per unit area coefficient was the ratio of two force per unit area difference ( below shows manometer, fig 1.4 )

Fig 1.4

This shows that we can utilize the differences in the tallness of the liquid to cipher the force per unit area coefficient. In other words, this can be said as the force per unit area coefficient is the ratio of the differences in the tallness of the liquid degrees which being measured.

Below equation ( fig 1.5 ) satisfy above statement.

Fig 1.5

This equation can be besides used to finding the force per unit area coefficient ( fig 1.6 )

Fig 1.6

Pi – force per unit area at tapping I

P? – free watercourse force per unit area

? – Air denseness

V – Free watercourse speed

S – Wing country

c – Aerodynamic mean chord

r – The manometer reading

h – The tallness of the liquid in the manometer

– Pressure coefficient

## Procedures

This subdivision explains the processs followed for transporting out the experiment.

The air current tunnel is connected to a multi-tube manometer for mensurating the force per unit area distribution around the airfoil surfaces. First, the manometer is adjusted to a 40o slope which is the optimal place for taking the readings.

Second, all the connexions between the air current tunnel and the manometer have to be checked so that it is ensured that the force per unit area is measured right. The higher the force per unit area on the airfoil, the lower the rise in liquid in the manometer will be and frailty versa.

Now, it is clip for puting the angle of incidence of the airfoil. This has to be done with the air current tunnel ‘s fan switched off as when it is switched on, the force per unit area exterior is higher than the force per unit area interior and the angle puting bosss would be difficult to revolve.

After the angle is set to -4o for the first portion of the experiment, the air current tunnel fan is switched on to 75 % of its maximal power. We so wait for one minute until all the liquid in all tubings is stable and so the readings are taken down, where the manometer measures the force per unit area in footings of tallness of the liquid. In add-on to the tubings mensurating the force per unit area over the airfoil at all tappings, there are two tubings for mensurating the stagnancy force per unit area, which may be assumed in this experiment to be the room force per unit area as it is an unfastened terminal air current tunnel, and the free watercourse force per unit area of the air current tunnel.

Finally after all the readings have been taken down, the air current tunnel fan is shut down and the above processs are repeated once more for mensurating the force per unit area distributions on the airfoil surface with angles of incidence 0o, 4o, 8o, 12o and 15o. The differences in highs of the liquid in the tubings are observed with regard to jumping angles and the stalling angle is worked out.

## Consequences

After the coefficient of force per unit area Cp was worked out utilizing the expression, graphs of CP versus tapping point place on the aerofoil ten with regard to the chord length degree Celsius i.e. , x/c, for every angle of incidence tested were plotted.

## /

## 1

## 2

## 3

## 4

## 5

## 6

## 7

## 8

## 9

## 10

## 11

## 12

## 13

## 14

## -4

21

28.6

33

38.2

49.4

53

49.6

42.8

33

44

47.4

49.4

51.6

52

## 0

14

45.8

48.4

52

58.4

59.8

53.8

43.8

29.6

38

39

37.8

35

33

## 4

38

64

63

64

65

63

52.4

43.2

27

33

32.8

29.2

24.2

21.4

## 8

63

71.5

70

69

65

58.8

46

42.2

27.8

31.6

30.2

25.6

20.2

17.6

## 12

87

80.4

76

72

62

49.5

43

43.2

28

30.2

27.8

22.4

17

14.8

## 15

93

88

76.5

70

57

46

43

43.4

26.6

28.4

25.4

20.4

15.4

13.8

The tabular array above shows the force per unit area recorded at 14 different points on the aerofoil that was obtained in the lab.

## degree Celsiuss

## ten

## x/c

## 1

140

0

0

## 2

140

6

0.042857

## 3

140

9.5

0.067857

## 4

140

15

0.107143

## 5

140

38.5

0.275

## 6

140

63.5

0.453571

## 7

140

90

0.642857

## 8

140

116.5

0.832143

## 9

140

8

0.057143

## 10

140

12.5

0.089286

## 11

140

26.5

0.189286

## 12

140

52

0.371429

## 13

140

78

0.557143

## 14

140

103.5

0.739286

Table demoing places of tappings on the aerofoil

## hi-h?

## 1

## 2

## 3

## 4

## 5

## 6

## 7

## 8

## 9

## 10

## 11

## 12

## 13

## 14

-4

-13.8

-6.2

-1.8

3.4

14.6

18.2

14.8

8

-1.8

9.2

12.6

14.6

16.8

17.2

0

-19.6

12.2

14.8

18.4

24.8

26.2

20.2

10.2

-4

4.4

5.4

4.2

1.4

-0.6

4

5.8

31.8

30.8

31.8

32.8

30.8

20.2

11

-5.2

0.8

0.6

-3

-8

-10.8

8

32

40.5

39

38

34

27.8

15

11.2

-3.2

0.6

-0.8

-5.4

-10.8

-13.4

12

57

50.4

46

42

32

19.5

13

13.2

-2

0.2

-2.2

-7.6

-13

-15.2

15

65

60

48.5

42

29

18

15

15.4

-1.4

0.4

-2.6

-7.6

-12.6

-14.2

Table demoing difference between gauge force per unit area and stagnancy force per unit area

## Cp

## 1

## 2

## 3

## 4

## 5

## 6

## 7

## 8

## 9

## 10

## 11

## 12

## 13

## 14

-4

0.6330

0.2844

0.0826

-0.1560

-0.6697

-0.8349

-0.6789

-0.3670

0.0826

-0.4220

-0.5780

-0.6697

-0.7706

-0.7890

0

0.9515

-0.5922

-0.7184

-0.8932

-1.2039

-1.2718

-0.9806

-0.4951

0.1942

-0.2136

-0.2621

-0.2039

-0.0680

0.0291

4

-0.3021

-1.6563

-1.6042

-1.6563

-1.7083

-1.6042

-1.0521

-0.5729

0.2708

-0.0417

-0.0312

0.1563

0.4167

0.5625

8

-1.7778

-2.2500

-2.1667

-2.1111

-1.8889

-1.5444

-0.8333

-0.6222

0.1778

-0.0333

0.0444

0.3000

0.6000

0.7444

12

-3.3529

-2.9647

-2.7059

-2.4706

-1.8824

-1.1471

-0.7647

-0.7765

0.1176

-0.0118

0.1294

0.4471

0.7647

0.8941

15

-4.3333

-4.0000

-3.2333

-2.8000

-1.9333

-1.2000

-1.0000

-1.0267

0.0933

-0.0267

0.1733

0.5067

0.8400

0.9467

Table demoing different values of coefficient of force per unit area Cp calculated utilizing expression

After all values were recorded and computations have been made, the graphs were ready to be plotted and they were as follows:

The 6 graphs above show the force per unit area distribution over the aerofoil at different angles of incidence. On each graph the force per unit area on the upper surface is marked in squares and on the lower surface is marked in diamonds.

After theses graphs have been plotted, it was clip for the country under the curve for every graph to be calculated therefore working out the coefficient of lift CL. The computation of the coefficient of lift can demo us easy where the aerofoil procrastinating angle is ; which is the angle at which the coefficient of lift is the upper limit.

## Calculating country under the graph

To cipher the country of the graph, several trigons were used. To happen an country of a trigon this expression was used

For illustration for trigon Angstrom:

A= ? x Base x Height

A=1/2 x 0.04 tens 0.24

A=0.0048

Rest of the computations for this graph was done in similar manner.

Triangle A

## Base

0.04

Area

## 0.0048

Triangle B

## Base

0.04

Area

## 0.0052

Triangle C

## Base

0.11

Area

## 0.0143

Triangle D

## Base

0.11

Area

## 0.0088

Triangle E

## Base

0.05

Area

## 0.004

Triangle F

## Base

0.26

Area

## 0.0078

Triangle G

## Base

0.24

Area

## 0.01356

Triangle H

## Base

0.1

Area

## 0.005

Triangle I

## Base

## Height

0.1

0.1

Area

## 0.005

Triangle J

## Height

## Base

0.3774

0.224

Area

## 0.042269

Entire Area = 0.0048+0.0052+0.0143+0.0088+0.004+0.0078+0.01356+0.005+0.005+0.042269

## Entire Area= 0.1107288

To happen the country between the graphs for this peculiar graph, several trigon, trapezium and rectangle were used.

A ( Triangle )

ten

Yttrium

0.1

0.51

Area

## 0.0255

F ( Triangle )

ten

Y

0.04

0.8

Area

## 0.016

To happen the country of a rectangle:

Area = a ten B

Area=0.06×0.3

Area=0.018

C ( Rectangle )

ten

Y

0.06

0.3

Area

## 0.018

Following computation were made to happen the country of a trapezium:

Therefore for D ( Trapezium ) :

h=0.2

Similarly in the same manner E ( Trapezium ) was calculated.

D ( Trapezium )

h=0.2

## a

## B

0.05

0.04

Area

## 0.009

E ( Trapezium )

h=0.2

## a

## B

0.04

0.03

Area

## 0.007

Trapezium Rule:

To cipher B ( Trapezium ) :

H = 0.04

Area = 0.04/2 ( ( 0.51+0.26 ) + 2 ( 0.5+0.49+0.47+0.45+0.41+0.39+0.36+0.32+0.3+0.29+0.26+0.25+0.24+0.24+0.23 )

## Area= 0.2234

To cipher G ( Trapezium ) :

h= 0.04

Area = 0.04/2 ( ( 0 + 0 ) + 2 ( 0.1+0.12+0.22+0.33+0.42+0.52+0.6+0.66+0.72+0.78+0.79+0.78+0.76+0.72+0.69+0.61+0.55+0.46+0.39+0.2 )

## Area =0.4168

## Entire Area = 0.0255+0.016+0.018+0.009+0.007+0.2234+0.4168

## Entire Area = 0.7157

To happen the country between the graphs, several trigons and trapeziums were used.

B ( Triangle )

## ten

## Y

0.09

1.37

Area

## 0.06165

D ( Triangle )

## ten

## Y

0.09

0.2

Area

## 0.009

E ( Triangle )

## ten

## Y

0.06

1.35

Area

## 0.0405

F ( Triangle )

## ten

## Y

0.06

0.58

Area

## 0.0174

G ( Triangle )

## ten

## Y

0.02

0.28

Area

## 0.0028

To cipher A ( Trapezium ) :

h= 0.04

Area = 0.04/2 ( ( 0 + 0.56 ) + 2 ( 0.54+0.5+0.48+0.42+0.39+0.33+0.29+0.22+0.18+0.12+0.09+0.05 )

## Area = 0.1556

To cipher C ( Trapezium ) :

h= 0.04

Area = 0.04/2 ( ( 0.81 + 1.62 ) + 2 ( 0.9+1+1.11+1.23+1.35+1.48+1.6+1.62+1.64+1.66+1.7+1.7+1.69+1.68 1.67+1.65 )

## Area = 0.9958

## Entire Area=0.06165+0.009+0.0405+0.0174+0.0028

## Entire Area=1.27375

B ( Triangle )

## Ten

0.05

Area =

## 0.018

To happen the country between the graphs, following computations were done:

A ( Triangle )

## Ten

0.58

Area=

## 0.2088

C ( Triangle )

## Ten

0.06

Area =

## 0.0204

E ( Triangle )

## Ten

0.06

Area

## 0.054

F ( Triangle )

## Ten

## Yttrium

0.06

0.45

Area

## 0.0135

To cipher the D ( Trapezium )

h= 0.04

Area = 0.04/2 ( ( 0.68 + 2.1 ) + 2 ( 0.7+0.72+0.8+0.9+1.01+1.2+1.3+1.4+1.52+1.6+1.69+1.75+1.8+1.85 1.91+1.96+2 )

## Area = 0.9958

## Entire Area= 0.2088+0.018+0.0204+0.054+0.0135+0.9958

## Entire Area= 1.3347

To happen the country between the graphs, following computations were done:

A ( Triangle )

ten

Y

0.74

0.89

Area

## 0.3293

B ( Triangle )

ten

Y

0.04

0.89

Area

## 0.0178

D ( Triangle )

ten

Y

0.04

2.8

Area

## 0.056

E ( Triangle )

ten

Y

0.04

0.65

Area

## 0.013

F ( Trapezium )

## a

## B

0.09

0.04

Area

## 0.04875

To cipher the C ( Trapezium )

h= 0.04

Area = 0.04/2 ( ( 0.75 + 2.8 ) + 2 ( 0.73+0.75+0.8+0.85+0.91+1.01+1.11+1.25+1.4+1.6+1.8+1.9+2.1+2.2 2.35+2.5 )

## Area = 1.0014

## Entire Area= 0.3293+0.0178+0.056+0.013+0.04875+1.0014

## Entire Area=1.46625

To happen the country between the graph following computations were done:

A ( Triangle )

ten

Y

0.64

0.96

Area

## 0.3072

B ( Triangle )

ten

Y

0.04

0.96

Area

## 0.0192

C ( Triangle )

ten

Y

0.04

1

Area

## 0.02

E ( Triangle )

ten

Y

0.04

3.4

Area

## 0.068

F ( Triangle )

ten

Y

0.04

0.9

Area

## 0.018

To cipher D ( Trapezium ) :

h= 0.04

Area = 0.04/2 ( ( 1 + 3.4 ) + 2 ( 1+1+1+1+1.02+1.04+1.1+1.18+1.3+1.42+1.6+1.8+2+2.2+2.4+2.6+2.8 )

## Area = 1.1464

## Entire Area= 0.3072+0.0192+0.02+0.068+0.018+1.1464

## Entire Area=1.5788

After the coefficient of lift CL was calculated for the different angles of incidence, a graph of CL versus angle of incidence was plotted and was as follows:

Angle of Incidence

Lift

-4

0.110729

0

0.7157

4

1.27375

8

1.3347

12

1.46625

15

1.5788

## Discussion

For the first set of graphs of Cp versus x/c obtained, the forms are right. If an aerofoil has a positive angle of onslaught it means it is mounting therefore the force per unit area distribution on the top surface of the aerofoil must be less than the force per unit area distribution on the lower surface. When the angle of incidence had a positive value, the force per unit area distribution on the lower surface was ever higher than the upper surface. Visually this can be seen clearly on the graphs as the squared curve was ever above the diamond curve significance force per unit area was higher on the lower surface.

However, when the angle of incidence had a negative value the scenario was wholly reversed intending the force per unit area distribution on the upper surface became higher than the force per unit area distribution on the lower surface therefore the aerofoil is in a province of falling. This can besides be seen visually from the graph with an angle of incidence of -4o as the squared line this clip was below the diamond line intending a higher force per unit area on the upper surface than on the lower surface.

After that, the coefficient of lift on the aerofoil for the different angles of incidence was calculated via the country under the curve of the first set of graphs. A graph of the coefficient of lift versus angle of incidence was so plotted and it merely complemented the predating consequences. At -4o, the lowest coefficient of lift was attained. As the angle was increased bit by bit the coefficient of lift kept lifting until it reached a maximal value at 15o which is the procrastinating angle. In the terminal, the values complement the consequences as they have proven that the maximal coefficient of lift was achieved at 15o which is the procrastinating angle.

## Decision

From the experiment, it can be concluded that in an aircraft, lift is caused by an upward force that is resulted from the difference in force per unit area between the top and the bottom surface of the wings. This difference in force per unit area is due to the design of the aerofoil, and the sum of the lift is dependent on the angle at which the wing is inclined.

In add-on, the force per unit area distribution over an aerofoil alterations with changing angles of incidence and the coefficient of lift can be calculated from a Cp versus angle of incidence graph. However, the coefficient of lift can be besides alterations with changing angles of incidence. This shows, the maximal coefficient of lift is achieved at the procrastinating angle.

However, in malice that the theory and the computations followed were right, two mistakes have been made. First, in many instances the force per unit area tallness was unstable, intending it was playing between many values so an mean value was taken which may hold caused in bend some mistakes in computations. Another mistake was that an angle higher than 15o should hold been tested so that it can be seen that the coefficient of lift would drop at any angle above 15o therefore corroborating that 15o is the procrastinating angle.

In the nutshell, by understanding the whole construct that has been explained above, we will be able to utilize it to cover with existent life state of affairss. That is, if we are able to cognize the forces that are exerted on an aerofoil, we will be able to plan aerofoil to cut down the forces such as drag force. By making so, the plane will be able to raise off ( take-off ) swimmingly at short distance. That nevertheless improves the fuel efficiency because if the retarding force force reduces it for illustration from 100 per cent to 45 per cent, it will let the aircraft to utilize less fuel when compared with the one which has no retarding force decrease.